Question

In each of the following problems you are given a function on the interval \(-\pi

Short answer

The Fourier series is: \( f(x) = \frac{1}{2} + \sum_{n=1}^{\infty} \frac{(-1)^n - 1}{n\pi} \sin(nx) \)

Step-by-step solution

- Understanding the Function

The given piecewise function is defined as follows:\( f(x) = \begin{cases} 1, & -\pi < x < 0 \ 0, & 0 < x < \pi \end{cases} \)This function is defined on the interval \(-\pi < x < \pi\) and repeats every \(2\pi\).

- Sketching Several Periods

Sketch the function over its interval \(-\pi < x < \pi\) and then repeat the pattern for several periods. The function is 1 from \(-\pi < x < 0\) and 0 from \(0 < x < \pi\).Repeat this pattern to sketch over multiple periods: \([-3\pi, 3\pi]\).

- Fourier Series General Form

The general form of the Fourier series for a function \(f(x)\) with period \(2\pi\) is given by:\( f(x) = a_0 + \sum_{n=1}^{\infty} \left(a_n \cos(nx) + b_n \sin(nx)\right) \)where:\( a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx \)\( a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \)\( b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \)

- Calculate \(a_0\)

Calculate \(a_0\):\( a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx \)Splitting the integral into two parts based on the function definition:\( a_0 = \frac{1}{2\pi} \left( \int_{-\pi}^{0} 1 \, dx + \int_{0}^{\pi} 0 \, dx \right) \)\( a_0 = \frac{1}{2\pi} \left( x \Bigg|_{-\pi}^{0} + 0 \right) \)\( a_0 = \frac{1}{2\pi} (0 - (-\pi)) = \frac{\pi}{2\pi} = \frac{1}{2} \)

- Calculate \(a_n\)

Calculate \(a_n\):\( a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \)Splitting the integral similarly:\( a_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} 1 \cos(nx) \, dx + \int_{0}^{\pi} 0 \cos(nx) \, dx \right) \)\( a_n = \frac{1}{\pi} \int_{-\pi}^{0} \cos(nx) \, dx \)\( a_n = \frac{1}{\pi} \left( \frac{\sin(nx)}{n} \Bigg|_{-\pi}^{0} \right) \)\( a_n = \frac{1}{\pi} \left( \frac{\sin(0)}{n} - \frac{\sin(-n\pi)}{n} \right) \)\( a_n = \frac{1}{\pi} \left( 0 - \frac{\sin(n\pi)}{n} \right) \)Since \(\sin(n\pi) = 0\) for any integer n:\( a_n = 0 \)

- Calculate \(b_n\)

Calculate \(b_n\):\( b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \)Splitting the integral similarly:\( b_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} 1 \sin(nx) \, dx + \int_{0}^{\pi} 0 \sin(nx) \, dx \right) \)\( b_n = \frac{1}{\pi} \int_{-\pi}^{0} \sin(nx) \, dx \)\( b_n = \frac{1}{\pi} \left( -\frac{\cos(nx)}{n} \Bigg|_{-\pi}^{0} \right) \)\( b_n = \frac{1}{\pi} \left( -\frac{\cos(0)}{n} - \frac{-\cos(-n\pi)}{n} \right) \)\( b_n = \frac{1}{\pi} \left( -\frac{1}{n} - \frac{\cos(n\pi)}{n} \right) \)Since \(\cos(n\pi) = (-1)^n\), we get:\( b_n = \frac{1}{\pi} \left( -\frac{1}{n} + \frac{(-1)^n}{n} \right) \)\( b_n = \frac{1}{\pi} \frac{(-1)^n - 1}{n} \)

- Write the Fourier Series

Combine the results to write the Fourier Series:\( f(x) = \frac{1}{2} + \sum_{n=1}^{\infty} \left(0 \cos(nx) + \frac{(-1)^n - 1}{n\pi} \sin(nx)\right) \)\( f(x) = \frac{1}{2} + \sum_{n=1}^{\infty} \frac{(-1)^n - 1}{n\pi} \sin(nx) \)

Key concepts

Periodic Functions

In mathematics, periodic functions are functions that repeat their values at regular intervals or periods. The concept of periodicity is fundamental in various fields, including engineering, physics, and signal processing. For any periodic function, there exists a period \( T \) such that for every value of \( x \) in the domain of the function, the following holds:

• \( f(x) = f(x + T) \)

A classic example is the sine function, \( \sin(x) \), which repeats every \( 2\pi \) units. In our exercise, the given function repeats itself every \( 2\pi \). It is defined piecewise within the interval \( -\pi < x < \pi \) and then continues to repeat this pattern indefinitely.
Sketching several periods of a periodic function helps in visualizing its repetitive nature, aiding in understanding its Fourier series representation. For instance, if you sketch the given function over multiple periods like \( [-3\pi, 3\pi] \), you'll see the piecewise segments alternate between 1 and 0, providing a clear visual comprehension of its periodic behavior.

Piecewise Functions

A piecewise function is a function that is defined by different equations over different intervals of its domain. Each piece of the function applies to a specific section of the domain. These functions are particularly useful for modeling real-world situations where a single formula is not sufficient to describe the entire behavior of the function.
In our given problem, the function \( f(x) \) is defined as:

• \( f(x) = 1 \) for \( -\pi < x < 0 \)
• \( f(x) = 0 \) for \( 0 < x < \pi \)

This means that the function has two distinct behaviors depending on the value of \( x \). It is 1 in the first half and 0 in the second half of one period. When handling piecewise functions, you usually analyze each piece separately to understand the overall behavior of the function. In this way, we can find the Fourier series by considering each section of the function.

Fourier Coefficients

The Fourier series of a periodic function is a way to represent the function as an infinite sum of sines and cosines. The coefficients in this series are known as Fourier coefficients. They can be calculated using integrals over one period of the function. The Fourier series has the general form:

• \( f(x) = a_0 + \sum_{n=1}^{\infty} \left(a_n \cos(nx) + b_n \sin(nx)\right) \)

Here are the steps to find the Fourier coefficients for our function:

• Calculate \( a_0 \) : This is the average value of the function over one period.
  
  \( a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx \)


• Calculate \( a_n \) : These coefficients are associated with the cosine terms.
  
  \( a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \)


• Calculate \( b_n \) : These coefficients are for the sine terms.
  
  \( b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \)

In our solution, after performing these calculations, we find:

• \( a_0 = \frac{1}{2} \)
• \( a_n = 0 \)
• \( b_n = \frac{1}{\pi} \frac{ (-1)^n - 1 }{ n } \)

The resulting Fourier series of our piecewise, periodic function can then be expressed by plugging these coefficients back into the general form.
Understanding and calculating these coefficients gives a deeper insight into the frequency components that make up the original function.